Probability Random Number Generator

Question

Let's say I'm writing a simple luck game - each player presses Enter and the game assign him a random number between 1-6. Just like a cube. At the end of the game, the player with the highest number wins.

Now, let's say I'm a cheater. I want to write the game so player #1 (which will be me) has a probability of 90% to get six, and 2% to get each of the rest numbers (1, 2, 3, 4, 5).

So, how can I generate a number random, and set the probability for each number?

Thanks.

Answer 1

static Random random = new Random();

static int CheatToWin()
{
    if (random.NextDouble() < 0.9)
        return 6;

    return random.Next(1, 6);
}

Another customizable way to cheat:

static int IfYouAintCheatinYouAintTryin()
{
    List<Tuple<double, int>> iAlwaysWin = new List<Tuple<double, int>>();
    iAlwaysWin.Add(new Tuple<double, int>(0.02, 1));
    iAlwaysWin.Add(new Tuple<double, int>(0.04, 2));
    iAlwaysWin.Add(new Tuple<double, int>(0.06, 3));
    iAlwaysWin.Add(new Tuple<double, int>(0.08, 4));
    iAlwaysWin.Add(new Tuple<double, int>(0.10, 5));
    iAlwaysWin.Add(new Tuple<double, int>(1.00, 6));

    double realRoll = random.NextDouble(); // same random object as before
    foreach (var cheater in iAlwaysWin)
    {
        if (cheater.Item1 > realRoll)
            return cheater.Item2;
    }

    return 6;
}

Answer 2

You have a few options, but one way would be to pull a number between 1 and 100, and use your weights to assign that to a dice face number.

So

1,2 = 1
3,4 = 2
5,6 = 3
7,8 = 4
9,10 = 5
11-100 = 6

this would give you the ratios you need, and would also be fairly easy to tune later.

Answer 3

you can define array of distribution (pseudocode) :

//fair distribution

array = {0.1666, 0.1666, 0.1666, 0.1666, 0.1666, 0.1666 };

then roll the dice from 0 to 1, save to x then do

float sum = 0;
for (int i = 0; i < 6;i++)
{
   sum += array[i];
   if (sum > x) break;
}

i is the dice number.

now if you want to cheat change array to:

array = {0.1, 0.1, 0.1, 0.1, 0.1, 0.5 };

and you will have 50% to get 6 (instead of 16%)